Question

这是我的第一个问题，我长期处理这个问题，但我不知道它在哪里。

我想从文件中读取一个Json字符串，（但确实如此）但是当我使用＆＃34; json_decode ＆＃34;不起作用，我试图将文件内容复制到一个变量中，并且效果很好，我不明白为什么，可以有人帮忙。

谢谢！

这里有代码：

<HTML>
<p>-JSON STRING (From File)-</p>
<?PHP
    $jsonStr = "";
    $jsonStr =$jsonStr . file_get_contents("./Config.json");
    echo "Json String: " . $jsonStr;

?>
<p>-JSON DATA-</p>
<?PHP
    $jsonDec = json_decode($jsonStr, true);
    $name = $jsonDec['config'][0]['nombre'];
    echo "Json data: " . $name;

    echo "</br>--------------------------------------------------------------------------------";
?>

<p>-JSON STRING (From var)-</p>
<?PHP
    $jsonStr = '{ "config" :[ {"name" : "Eduvi", "link" : "LINKURL"}, {"name" : "Eduvi", "link" : "LINKURL"}, {"name" : "Eduvi", "link" : "LINKURL"} ] } ';
    echo "Json String: " . $jsonStr;
?>
<p>-JSON DATA-</p>
<?PHP
    $jsonDec = json_decode($jsonStr, true);
    $name = $jsonDec['config'][0]['name'];

    echo "Json data: " .  $name;
?>

以下是我得到的结果：

-JSON STRING (From File)-

Json String：{＆＃34; config＆＃34; ：[{＆＃34; name＆＃34; ：＆＃34; Eduvi＆＃34;，＆＃34;链接＆＃34; ：＆＃34; LINKURL＆＃34;}，{＆＃34; name＆＃34; ：＆＃34; Eduvi＆＃34;，＆＃34;链接＆＃34; ：＆＃34; LINKURL＆＃34;}，{＆＃34; name＆＃34; ：＆＃34; Eduvi＆＃34;，＆＃34;链接＆＃34; ：＆＃34; LINKURL＆＃34;}]}

-JSON DATA- Json数据：

-JSON STRING（来自var） - Json String：{＆＃34; config＆＃34; ：[{＆＃34; name＆＃34; ：＆＃34; Eduvi＆＃34;，＆＃34;链接＆＃34; ：＆＃34; LINKURL＆＃34;}，{＆＃34; name＆＃34; ：＆＃34; Eduvi＆＃34;，＆＃34;链接＆＃34; ：＆＃34; LINKURL＆＃34;}，{＆＃34; name＆＃34; ：＆＃34; Eduvi＆＃34;，＆＃34;链接＆＃34; ：＆＃34; LINKURL＆＃34;}]}

-JSON DATA- Json数据：Eduvi

Answer 1

Undefined index: nombre in /tmp/test.php on line 10 =＆gt;意味着你调用一个不存在的索引，正如你自己在数据中看到的那样;如果是，它将显示包含的值

输入文件site.txt：

{ "config" :[ {"name" : "Eduvi", "link" : "LINKURL"}, {"name" : "Eduvi", "link" : "LINKURL"}, {"name" : "Eduvi", "link" : "LINKURL"} ] }

php code'test.php'：

<?php

    echo("\n-JSON STRING (From File)-\n");
    $jsonStr = "";
    $jsonStr = trim( file_get_contents("./site.txt") ); // fix 
    echo "Json String: " . $jsonStr."\n";

    echo("\n-JSON DATA-\n");
    $jsonDec = json_decode($jsonStr, true);
    $name = $jsonDec['config'][0]['nombre']; //index nombre does not exist, if it is 'name', then it works
    echo "Json data: " . $name;

    echo "\n--------------------------------------------------------------------------------\n";

    echo("\n-JSON STRING (From var)-\n");
    $jsonStr = '{ "config" :[ {"name" : "Eduvi", "link" : "LINKURL"}, {"name" : "Eduvi", "link" : "LINKURL"}, {"name" : "Eduvi", "link" : "LINKURL"} ] } ';
    echo "Json String: " . $jsonStr."\n";
    echo("\n-JSON DATA-\n");
    $jsonDec = json_decode($jsonStr, true);
    $name = $jsonDec['config'][0]['name'];

    echo "Json data: " .  $name."\n";
?>

使用php test.php

在cli上运行它

-JSON STRING (From File)-
Json String: { "config" :[ {"name" : "Eduvi", "link" : "LINKURL"}, {"name" : "Eduvi", "link" : "LINKURL"}, {"name" : "Eduvi", "link" : "LINKURL"} ] }

-JSON DATA-
PHP Notice:  Undefined index: nombre in /tmp/test.php on line 10
Json data:
--------------------------------------------------------------------------------

-JSON STRING (From var)-
Json String: { "config" :[ {"name" : "Eduvi", "link" : "LINKURL"}, {"name" : "Eduvi", "link" : "LINKURL"}, {"name" : "Eduvi", "link" : "LINKURL"} ] }

-JSON DATA-
Json data: Eduvi

修复'name'后输出

：

-JSON STRING (From File)-
Json String: { "config" :[ {"name" : "Eduvi", "link" : "LINKURL"}, {"name" : "Eduvi", "link" : "LINKURL"}, {"name" : "Eduvi", "link" : "LINKURL"} ] }

-JSON DATA-
Json data: Eduvi
--------------------------------------------------------------------------------

-JSON STRING (From var)-
Json String: { "config" :[ {"name" : "Eduvi", "link" : "LINKURL"}, {"name" : "Eduvi", "link" : "LINKURL"}, {"name" : "Eduvi", "link" : "LINKURL"} ] }

-JSON DATA-
Json data: Eduvi

也尝试使用与此处相同的输入进行测试;）

Answer 2

如果您的Config.json是以UTF-8（with BOM）编码的，则json_decode函数将失败并返回NULL。

您可以通过修剪$ jsonStr字符串中的BOM来解决此问题：

$jsonStr = trim(file_get_contents("config.json"), "\xEF\xBB\xBF");

除此之外，您还没有解决现有价值（请参阅vv01f＆＃39> 的答案）

祝你好运！

Answer 3

我正在测试这个例子，完全正常，基本文件是ob.php

$jsonStr = "";
$jsonStr =$jsonStr . file_get_contents("myfile.json");
echo "Json String one : " . $jsonStr;


$jsonDec = json_decode($jsonStr, true);
$name = $jsonDec['config'][0]['name'];
echo "Json data: " . $name;

echo "</br>-------------------------";

之后添加json文件myfile.json

{
    "config": [{
        "name": "Eduvi",
        "link": "LINKURL"
    }, {
        "name": "Eduvi",
        "link": "LINKURL"
    }, {
        "name": "Eduvi",
        "link": "LINKURL"
    }]
}

它显示name的值，即数组中的第一个对象

PHP Json String解码

3 个答案: